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        riddles >> medium >> Divide Two Sets Of  Quadratic, Get Integers
        
(Message started by: K Sengupta on Jul 24^th, 2007, 8:01am)

Title: Divide Two Sets Of Quadratic, Get Integers
Post by K Sengupta on Jul 24^th, 2007, 8:01am

Analytically determine all possible pairs of positive integers (a, b) such that :

(a + b²)/(a²-b) and (a²+b)/(b²-a) are both positive integers.

Title: Re: Divide Two Sets Of Quadratic, Get Intege
Post by Eigenray on Jul 26^th, 2007, 2:41pm

[hideb]If a=b, then (a+a²)/(a²-a) = (a+1)/(a-1) = 1 + 2/(a-1), so a-1 | 2, so we have a=b=2 or a=b=3.

Otherwise, WLOG suppose a<b.

(a²+b)/(b²-a) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 1, so if a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif (b-2), then we would have
b²-(b-2) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif b²-a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif a²+b http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif (b-2)²+b,
or 2b http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 2, giving b=1 and a<0, a contradiction.

So we must have a=b-1. So (b²+b-1)/(b²-3b+1) is an integer. Since it's clearly > 1, it must be at least 2, so
b²+b-1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif 2(b²-3b+1), or
b²-7b+3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif 0,
which gives b < 7. Checking b=3,4,5,6, the only solution is (a,b) = (2,3).[/hideb]